Matrices and Determinants

We must have $a^2-b^2=1 < p$

$ (a+b)(a-b)=1 < p $

Either $a-b=0$ or $a+b$ is a multiple of $p$ when $a=b$ number of matrices is $p$ and when $a+b$ $=$ multiple of $p$.

$ \Rightarrow a, b \text { has } p-1 $

$ \begin{aligned} \text { Total number of matrices } & =p+p-1 \\\\ & =2 p-1 \end{aligned} $

We have an odd prime $p$. Consider the set

$ T_p \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b \\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of such matrices $A$ that satisfy:

$\text{trace}(A) \not\equiv 0 \pmod{p}$.

Since $\text{trace}(A) = a + a = 2a$, this means

$ 2a \;\not\equiv\; 0 \pmod{p}. $

Because $p$ is an odd prime, $2$ is invertible $\bmod\,p$.

Hence $2a \equiv 0 \pmod{p}$ if and only if $a \equiv 0 \pmod{p}$.

Thus the condition

$\text{trace}(A) \not\equiv 0 \pmod{p}$

is equivalent to

$ a \;\neq\; 0 \pmod{p}. $

In other words, $a$ must be a nonzero element modulo $p$; so

$a\in\{1,2,\dots,p-1\}.$

$\det(A) \equiv 0 \pmod{p}$.

For the matrix

$\begin{pmatrix} a & b \\ c & a \end{pmatrix},$ we have

$ \det(A) \;=\; a^2 - bc. $

The condition $\det(A)\equiv 0\pmod{p}$ means

$ a^2 \;-\; bc \;\equiv\; 0 \pmod{p} \quad\Longleftrightarrow\quad bc \;\equiv\; a^2 \pmod{p}. $

Putting these two pieces together:

$a$ is nonzero ($a \in \{1,\dots,p-1\}$).

$bc \equiv a^2 \pmod{p}.$

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Counting the solutions

We must count the number of triples $(a,b,c)$ with $a\neq 0$ and $bc \equiv a^2$.

Range for $a$

Since $a\in \{1,2,\dots,p-1\}$, there are $p-1$ possible values of $a$.

Condition $bc \equiv a^2$ for given $a$

Since $a \neq 0$, $a^2$ is also nonzero $\bmod\,p$.

If $b = 0$, then $bc \equiv 0$, which would only be equal to $a^2$ if $a^2 \equiv 0$. But $a^2\neq 0$ since $a\neq 0$. So $b=0$ gives no solutions.

Similarly, if $c=0$, then $bc=0$, which again cannot equal the nonzero $a^2$. So $c=0$ gives no solutions either.

Therefore $b$ and $c$ must both be nonzero modulo $p$.

Once $b$ is chosen to be any nonzero element in $\{1,\dots,p-1\}$, there is a unique $c \equiv a^2 \,b^{-1} \pmod{p}$. Hence:

For each nonzero $b$ (that is, $b\in\{1,\dots,p-1\}$), there is exactly one $c\in\{1,\dots,p-1\}$ satisfying $bc \equiv a^2$.

Consequently, for each fixed nonzero $a$, the number of $(b,c)$-pairs is $p-1$.

Putting it all together

We have $p-1$ choices for $a\neq 0$.

For each such $a$, there are $p-1$ pairs $(b,c)$ satisfying $bc\equiv a^2$.

Therefore, in total, the number of matrices is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

Hence the correct answer matches option C, which is

$ \boxed{(p-1)^2.} $

Let $p$ be an odd prime number, and consider the set

$ T_{p} \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b\\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of matrices $A \in T_p$ whose determinant is not divisible by $p$.

Recall that for

$ A \;=\; \begin{pmatrix} a & b \\ c & a \end{pmatrix}, $

the determinant is

$ \det(A) \;=\; a^2 \;-\; b\,c. $

Hence $\det(A)$ is not divisible by $p$ precisely when

$ a^2 \;-\; b\,c \;\not\equiv\; 0 \pmod{p}, \quad\text{i.e.,}\quad a^2 \;\neq\; b\,c \pmod{p}. $

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Total number of matrices

Since $a,b,c$ each range over $\{0,1,\dots,p-1\}$, there are

$ p^3 $

total possible matrices in $T_p$.

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Step 1: Count how many $(a,b,c)$ do satisfy $a^2 \equiv b\,c \pmod{p}$

It will be easier to first count the number of triples $(a,b,c)$ for which

$ a^2 \;\equiv\; b\,c \pmod{p}, $

and then subtract that from $p^3$.

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Case A: $a = 0$

Then $a^2 = 0$. We need

$ b\,c \;\equiv\; 0 \pmod{p}. $

Over the field $\mathbf{F}_p$, the product $b\,c\equiv 0$ if and only if at least one of $b$ or $c$ is $0$ (since $p$ is prime).

If $b=0$, then $c$ can be anything in $\{0,\dots,p-1\}$.

This gives $p$ possibilities.

If $c=0$, then $b$ can be anything in $\{0,\dots,p-1\}$.

This also gives $p$ possibilities.

However, the pair $(b=0, c=0)$ is counted in both cases, so we must subtract it out once.

Hence, for $a=0$, the number of $(b,c)$ such that $b\,c \equiv 0$ is

$ p \;+\; p \;-\; 1 \;=\; 2p \;-\; 1. $

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Case B: $a \neq 0$

Then $a$ can be any nonzero element in $\{1,2,\dots,p-1\}$; there are $p-1$ such choices. In that scenario, $a^2 \neq 0\pmod{p}$. We want

$ b\,c \;\equiv\; a^2 \pmod{p}. $

Over $\mathbf{F}_p$, a nonzero right-hand side $a^2$ can be written as $b\,c$ in the following way:

If $b=0$, then $b\,c = 0$, which cannot equal $a^2\neq 0$.

No solutions in that subcase.

If $b \neq 0$, then for each nonzero $b$ there is exactly one $c$ satisfying $b\,c \equiv a^2\pmod{p}$, namely $c \equiv a^2 b^{-1}\pmod{p}$.

Since $b$ can be any of the $p-1$ nonzero elements, we get $p-1$ solutions $(b,c)$ for each nonzero $a$.

Thus, for each $a\neq 0$, there are $p-1$ pairs $(b,c)$. Hence, for $p-1$ values of $a$, the total number of solutions in this case is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

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Total number of solutions to $a^2 = b\,c$

Summing these two cases:

Case A $(a=0)$: $2p - 1$ solutions.

Case B $(a\neq 0)$: $(p-1)^2$ solutions.

Hence the total count of $(a,b,c)$ satisfying $a^2 \equiv b\,c$ is

$ (2p -1) \;+\; (p-1)^2 \;=\; (2p -1) \;+\; \bigl(p^2 \;-\;2p \;+\;1\bigr) \;=\; p^2. $

(A classic result: there are exactly $p^2$ solutions to $b\,c = a^2$ over $\mathbf{F}_p^3$.)

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Step 2: Count how many $(a,b,c)$ satisfy $a^2 \neq b\,c$

Since there are $p^3$ total triples $(a,b,c)$, the number that satisfy

$ a^2 \;\neq\; b\,c \pmod{p} $

is simply

$ p^3 \;-\; p^2. $

This directly corresponds to matrices $A$ whose determinant $a^2 - b\,c$ is $\not\equiv 0\pmod{p}$.

Hence, the number of matrices in $T_p$ with $\det(A)$ not divisible by $p$ is

$ \boxed{p^3 - p^2}. $

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The expression $p^3 - p^2$ matches Option D.

Here,

Matrix [A]_{3 × 3} is symmetric

Here, 5 be the number of 1’s and 4 be the 0’s entry in matrix A

Let x be the number of 1’s on main diagonal.

Since A is symmetric so, y be the number of 1’s below the main diagonal

So , x + 2y = 5

x = 1 and y = 2

or x = 3 and y = 1

Case I : x = 1 and y = 2

Here, Main diagonal entries in 0, 0 and 1. Hence, we can choose main diagonal in 3 ways, and element above the diagonal is 1, 1, 0 . Hence, we can choose it element above the main diagonal in 3 ways , element of below diagonal depends on above the main diagonal.

Hence, total way = 3 × 3 = 9

Case II : x = 3 and y = 1

Here, Main diagonal entries in 1, 1 and 1.
Hence, we can choose main diagonal 1 ways and element above the diagonal is 1, 0, 0. Hence, we can choose it element above the main diagonal in 3 ways, element of below diagonal depends on above the main diagonal.

Hence, total ways = 3 × 4 = 12

So, Total matrices = 9 + 3 = 12

We have

$\left[ {\matrix{ 0 & a & b \cr a & 0 & c \cr b & c & 1 \cr } } \right]$

We can see that either $b = 0$ or $c = 0 \Rightarrow |A| \ne 0$; therefore two matrices.

$\left[ {\matrix{ 0 & a & b \cr a & 1 & c \cr b & c & 0 \cr } } \right]$

Now, either $a = 0$ or $c = 0 \Rightarrow |A| \ne 0$; therefore, two matrices

$\left[ {\matrix{ 1 & a & b \cr a & 0 & c \cr b & c & 0 \cr } } \right]$

Now, either $a = 0$ or $b = 0 \Rightarrow |A| \ne 0$; therefore two matrices.

$\left[ {\matrix{ 1 & a & b \cr a & 1 & c \cr b & c & 1 \cr } } \right]$

When $a = b = 0 \Rightarrow |A| = 0$

When $a = c = 0 \Rightarrow |A| = 0$

When $b = c = 0 \Rightarrow |A| = 0$

Therefore, there are only six matrices.

The six matrices A for which $|A| = 0$ are as follows:

$\left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 1 \cr 1 & 1 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 1 & 1 \cr 0 & 1 & 0 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 1 & 1 \cr 1 & 0 & 0 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow $ infinite solutions.

$\left[ {\matrix{ 1 & 1 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 1 & 1 \cr } } \right] \Rightarrow $ infinite solutions.

The given equations are

$x - 2y + 3z = - 1$

$ - x + y - 2z = k$

$x - 3y + 4z = 1$

$D = \left| {\matrix{ 1 & { - 2} & 3 \cr { - 1} & 1 & { - 2} \cr 1 & { - 3} & 4 \cr } } \right| = 0$

$D = \left| {\matrix{ 1 & { - 1} & 3 \cr { - 1} & k & { - 2} \cr 1 & 1 & 4 \cr } } \right| = k - 3 \ne 0$

If $k \ne 3$, the system has no solutions.

Hence, Statement - 1 is True Statement - 2 is correct explanation for Statement - 1.

$\begin{aligned} & A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right] \\ & \text { Let, } \quad \mathrm{U}_{1}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \\ & \mathrm{AU}_{1}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{c} x \\ 2 x+y \\ 3 x+2 y+z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow \quad x=1,2 x+y=0,3 x+2 y+z=0 \\ & y=-2,3-4+z=0 \\ & z=1 \text {. } \\ & \mathrm{U}_{1}=\left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] \end{aligned}$

$\begin{aligned} \text { Similarly } U_{2} & =\left[\begin{array}{c} 2 \\ -1 \\ -4 \end{array}\right] U_{3}=\left[\begin{array}{c} 2 \\ -1 \\ -3 \end{array}\right] \\ \therefore \quad U & =\left[\begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{array}\right] \\ |U| & =1(3-4)-2 \times 7+2 \times 5 \\ & =-15+18=3 \end{aligned}$

$ \mathrm{U}^{-1} \text { can be calculated by following shortcut } $

$ \begin{aligned} &\mathrm{U}^{-1}=\frac{1}{3}\left[\begin{array}{ccc} -1 & -2 & 0 \\ -7 & -5 & -3 \\ 9 & 6 & 3 \end{array}\right]\\ &\text { Sum of elements of } \mathrm{U}^{-1}=0 \end{aligned} $

$ \begin{aligned} \text { Solution } & =\left[\begin{array}{lll} 3 & 2 & 0 \end{array}\right] \cup\left[\begin{array}{l} 3 \\ 2 \\ 0 \end{array}\right] \\ & =\left[\begin{array}{lll} 3 & 2 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 0 \end{array}\right] \\ & =\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 0 \end{array}\right]=-3+8=5 \end{aligned} $